Dinitrogen pentoxide, N2O5, decomposes by first-order kinetics with a rate constant of 0.15 s-1 at 353 K.

(a) What is the half-life (in seconds) for the decomposition of N2O5 at 353 K?

(b) If [N2O5]0 = 0.0579 mol·L-1, what will be the concentration of N2O5 after 4.0 s?

(c) How much time (in minutes) will elapse before the N2O5 concentration decreases from 0.0579 mol·L-1 to 0.0143 mol·L-1?

To solve this problem, we can use the integrated rate law for a first-order reaction:

ln([A]/[A]0) = -kt

where [A] is the concentration at a given time, [A]0 is the initial concentration, k is the rate constant, and t is the time.

(a) To find the half-life, we can rearrange the integrated rate law equation as follows:

ln(1/2) = -kt1/2

The half-life (t1/2) is the time it takes for the concentration to decrease by half.

Substituting the given values into the equation:

ln(1/2) = -0.15 s-1 * t1/2

ln(1/2) = -0.15 t1/2

Now, we can solve for t1/2 using natural logarithms:

t1/2 = ln(1/2) / -0.15
t1/2 ≈ 4.61 seconds

Therefore, the half-life for the decomposition of N2O5 at 353 K is approximately 4.61 seconds.

(b) To find the concentration of N2O5 after 4.0 seconds, we can use the integrated rate law again:

ln([N2O5]/[N2O5]0) = -kt

Rearranging the equation:

[N2O5] = [N2O5]0 * e^(-kt)

Substituting the given values:

[N2O5] = 0.0579 mol·L-1 * e^(-0.15 s-1 * 4.0 s)

[N2O5] ≈ 0.0415 mol·L-1

Therefore, the concentration of N2O5 after 4.0 seconds is approximately 0.0415 mol·L-1.

(c) To find the time it takes for the N2O5 concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1, we can rearrange the equation and solve for t:

ln([N2O5]/[N2O5]0) = -kt

t = ln([N2O5]/[N2O5]0) / -k

Substituting the given values:

t = ln(0.0143 mol·L-1 / 0.0579 mol·L-1) / -0.15 s-1

t ≈ 8.86 seconds

Since there are 60 seconds in a minute, we can convert the time from seconds to minutes:

t = 8.86 s * (1 min / 60 s)

t ≈ 0.148 minutes

Therefore, it takes approximately 0.148 minutes for the N2O5 concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1.

To solve these problems, we can use the first-order kinetics equation:

ln([N2O5]t/[N2O5]0) = -kt,

where [N2O5]t is the concentration of N2O5 at time t, [N2O5]0 is the initial concentration, k is the rate constant, and t is the time.

(a) The half-life (t1/2) is the time taken for the concentration to decrease to half of the initial concentration ([N2O5]0/2). We can use the formula for half-life of a first-order reaction:

t1/2 = ln(2)/k.

Substituting the given value of k = 0.15 s-1, we can calculate:

t1/2 = ln(2)/0.15 ≈ 4.6229 s.

Therefore, the half-life of the decomposition of N2O5 at 353 K is approximately 4.6229 seconds.

(b) To calculate the concentration of N2O5 after 4.0 seconds, we can rearrange the first-order kinetics equation:

[N2O5]t = [N2O5]0 * e^(-kt).

Substituting the given values [N2O5]0 = 0.0579 mol·L-1, k = 0.15 s-1, and t = 4.0 s:

[N2O5]t = 0.0579 * e^(-0.15*4) ≈ 0.0452 mol·L-1.

Therefore, the concentration of N2O5 after 4.0 seconds is approximately 0.0452 mol·L-1.

(c) To find the time it takes for the concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1, we can rearrange the first-order kinetics equation:

t = -(ln([N2O5]t/[N2O5]0))/k.

Substituting the given values [N2O5]0 = 0.0579 mol·L-1, [N2O5]t = 0.0143 mol·L-1, and k = 0.15 s-1:

t = -(ln(0.0143/0.0579))/0.15 ≈ 14.9 s.

Since we want the time in minutes, we divide by 60:

t = 14.9/60 ≈ 0.2483 min.

Therefore, it will take approximately 0.2483 minutes for the N2O5 concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1.

a. k = 0.693/t1/2

b. ln(No/N) = kt.
No = given
N =?
k = from part a.
t = 4 s.

c.
same equation as b but
No = given
N = given
k = from a
t = ?
Note that part a is given in seconds, part c wants the answer in min.